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Problem at hand: Let $f(x)$ be a measurable function on a measure space $(\Omega, \sum, \mu)$. Then there exists a sequence of simple functions (on $\Omega$), uniformly converging to $f$.

My definition of simple function is any measurable function with a countable range.

The sequence used in the proof is defined by

$$f_n(x)=\frac{m}{n} \quad \text{ if }\, \frac{m}{n} \leq f(x) < \frac{m+1}{n}$$ ($n \in \mathbb{N}$ and $m \in \mathbb{Z}$) And can be shown to be measurable and has a countable range.

Does it truly converge uniformly to $f$? I've seen many questions discussing this but with equivalent definitions of simple functions and more assumptions. For example one uses finite range in the definition of simple function and then require the original function $f$ to be bounded in order to ensure uniform convergence, otherwise the convergence is only pointwise.

So, with my definition and particular sequence, do I need boundedness to ensure uniform convergence? Sorry for the possible duplicate, but I wanted to specifically ask with my definitions and this sequence (the others I saw use $m/2^j$ type sequences and are specifically finding increasing sequences converging to non-negative measurable functions $f$, here i'm not assuming anything on $f$ besides being measurable). Thanks in advanced for any clarifications.

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Let $E_m = \{m/n\le f< (m+1)/n\}, m \in \mathbb Z.$ Then

$$0\le f - \sum_{m\in \mathbb Z}\frac{m}{n}\chi_{E_m} < \frac{1}{n}$$

everywhere.

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  • $\begingroup$ This is what I did in my work and I thought it was fine but i was confused about seeing other variations of the same problem. so is it only because other definitions of simple function use finite ranges (or finite sums involving characteristic functions) that the convergence cant be uniform everywhere (say, when $f$ is unbounded)? $\endgroup$ Dec 16, 2015 at 20:36
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    $\begingroup$ Right, it's not possible to uniformly approximate an unbounded function by bounded functions. $\endgroup$
    – zhw.
    Dec 16, 2015 at 21:04

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